How to solve this Boundary Value problem?

Question

$$ \left(\frac{d^2 u}{d x^2} \right)+\frac{2}{x+x_0} \frac{d u}{d x}=\frac{\lambda}{k} u; \quad x \in [0,L], k,x_0>0 $$

$$u(0)=0$$ $$u(L)=0$$

Answer 1

Rewrite

$$ (x+x_0)^2\frac{d^2u}{dx^2} + 2(x+x_0)\frac{du}{dx} - \frac{\lambda}{k}(x+x_0)^2u = 0 $$

A non-trivial solution only exists for $\lambda < 0$. The substitution $z = \sqrt{\frac{-\lambda}{k}}(x+x_0)$ gives

$$ \frac{d^2u}{dz^2} + 2z\frac{du}{dz} + z^2 = 0 $$

The solution is in the form of spherical Bessel functions for $n=0$ which happen to have a closed form

$$ u(z) = c_1j_0(z) + c_2y_0(z) = \frac{c_1\sin z - c_2\cos z}{z} $$

The boundary conditions are $u(\alpha x_0) = u(\alpha (L+x_0)) = 0$ where $\alpha = \sqrt{\frac{-\lambda}{k}}$. We need to pick $\alpha$ such that two of the zeroes of this function align with the boundary.

Case 1: If $c_2=0$, then the zeroes are the zeroes of $\sin z$, which are $z_n = n\pi$. Since any 2 roots always differ by an integer multiple of $\pi$, we have

$$ \alpha(L+x_0) - \alpha x_0 = \alpha L = m \pi \implies \alpha = \frac{m\pi}{L} $$

where $m$ is an integer. Note that $\alpha x_0 = \frac{m\pi x_0}{L}$ must also be an integer multiple of $\pi$, therefore $\frac{x_0}{L}$ must be rational.

Case 2: If $c_1=0$, then the zeroes are the zeroes of $\cos z$, which are $z_n = (n+\frac12)\pi$. Similarly, these roots also differ by an integer multiple of $\pi$, so we have the same eigenvalue as Case 1, with the condition

$$ \alpha x_0 = \frac{m\pi x_0}{L} = (2n+1)\frac{\pi}{2} $$

or $\frac{2x_0}{L} = \frac{2n+1}{m}$ must be rational.

Case 3: If $c_1,c_2 \ne 0$, then we can write

$$ u(z) = \frac{c_1(\sin z - \mu \cos z)}{z} $$

where the zeroes occur at $\tan z_n = \mu$ for some $\mu \in (0,\infty)$. These roots also differ by an integer multiple of $\pi$, so yet again we have $\alpha = \frac{m\pi}{L}$.

Combining the 3 cases, the eigenfunction can be simplified as

$$ u_m(z) = \frac{\cos z_m \sin z - \sin z_m \cos z}{z} $$

where $z_m = \alpha_m x_0 = \frac{m\pi x_0}{L}$